1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Evaluation of Integral (nln(n))^-1

  1. Aug 4, 2012 #1
    1. The problem statement, all variables and given/known data

    Evaluate [tex]\int{\frac{1}{nln(n)} dn }[/tex]

    2. Relevant equations



    3. The attempt at a solution

    I know the answer thanks to WolframAlpha, I just want to understand why my method didn't work.

    I took a stab at parts using:
    [tex]u=\frac{1}{ln(n)}[/tex]
    [tex]\frac{dv}{dn}=\frac{1}{n}[/tex]
    So this gives:
    [tex]v=ln(n)[/tex]
    Using quotient rule:
    [tex]\frac{du}{dn}=\frac{0-(1)(1/n)}{(ln(n))^2}=\frac{-1}{n(ln(n))^2}[/tex]
    And therefore as:
    [tex]\int{u\frac{dv}{dn} dn } = uv - \int{v\frac{du}{dn} dn}[/tex]
    [tex]\int{\frac{1}{nln(n)} dn } = ln(n)\frac{1}{ln(n)} - \int{ln(n)\frac{-1}{n(ln(n))^2} dn}[/tex]
    [tex]= 1 + \int{\frac{1}{n ln(n)} dn}[/tex]

    Which surely isn't right? Can anyone spot the mistake/reason why this method doesn't work?

    Cheers
     
  2. jcsd
  3. Aug 4, 2012 #2

    uart

    User Avatar
    Science Advisor

    Hi Jimbobian. Write the integral as,

    [tex]\int n^{-1} \, \left[ ln(n)\right]^{-1} dn [/tex]

    and then note that it is in the form of [itex]\int g' \, f \circ g(n) \, dn [/itex]

    This is a special form (the product of a function of a function times the derivative of the inner function) that we should always be on the look out for when trying to find an anti-derivative. This form always has a simple solution so long as we know the anti derivative of just the outer function "f()".
     
  4. Aug 4, 2012 #3
    Damn, one month after school finishing and I'd already forgotten to be on a look out for that, despite the number of times my teacher brought it up!

    That form would certainly make evaluating it easier (and thank you for spotting it), but I would also like to know why my method didn't work?
     
  5. Aug 4, 2012 #4

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    We pretty much always ignore the constants of integration when we do integration by parts. When you integrate dv we take the constant zero. Your result just says the two antiderivatives differ by a constant, but that doesn't get you anywhere.

    I assume you know that integral would be done with a u substitution, letting u = ln(x).
     
  6. Aug 4, 2012 #5
    Thank you! Yes, the actual integration, as uart pointed out, isn't challenging I just got hung up on why this didn't work.

    Thank you both, my issue is solved!
     
  7. Aug 4, 2012 #6

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    If you do a definite integration instead, say from 2 to x, integration by parts just gives you
    [tex] \int_2^x \frac{dn}{n \ln(n)} = \int_2^x \frac{dn}{n \ln(n)},[/tex]
    which is no longer contradictory, but is also not useful.

    RGV
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Evaluation of Integral (nln(n))^-1
  1. Evaluating integral (Replies: 7)

  2. Evaluate the Integral (Replies: 5)

  3. Evaluate this integral (Replies: 4)

  4. Evaluate the integral (Replies: 7)

  5. Integral evaluation (Replies: 5)

Loading...